T. J. Dolan Magnetic field calculations Coil forces Coil forces RLC - - PowerPoint PPT Presentation

• Ch. 3. Pulsed and Water‐Cooled Magnets
• T. J. Dolan

Magnetic field calculations Coil forces Coil forces RLC circuit equations Distribution of J and B E t Energy storage Switching and transmission Magnetic flux compression Component reliability Power and cooling requirements Coil design considerations

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Coil design considerations

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Background

Chinese Discovered magnetism and invented compass. Mapped the world using compass and astronomical navigation. Zheng He brought knowledge to Europe in 1434 Oersted deflection of compass by current in wire Ampere interaction of current carrying wires Ampere interaction of current-carrying wires Faraday magnetic induction Maxwell equations of electromagnetism

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Required magnetic field equ ed ag e c e d

T = 2x108 K, n = 2x1020 m-3,  = 0.1  B = 5.9 T Field at coil is larger: Bcoil = Bo(Ro/Rcoil)

coil

• (
• coil)

Ro = 6 m, Rcoil = 2.5 m  B

il = 14 T

Rcoil R Bcoil 14 T Ro

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Advantages of Water-Cooled Magnets Advantages of Water Cooled Magnets

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(eh and fg are equal and opposite.)

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Toroidal Magnetic Field Ripple Toroidal Magnetic Field Ripple

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Law of Biot-Savart

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Field from a Circular Current Ring Field from a Circular Current Ring

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k2 K(k) E(k) k2 K(k) E(k) k2 K(k) E(k) k2 K(k) E(k)

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Example – Field of Circular Coil

Circular coil, a = 0.5 m, I = 100 kA. Find B( r = 0.4 m, z = 0.6 m ) k2 = 0.683707 K(k) = 2.05215 E(k) = 1.25125 Br = 0.0025 T Bz = 0.027 T

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Coil Forces

F =  dF =  JxB dV F =  dF =  JxB dV F =  dF = I  dℓxB for thin wires wires B1 (at I2) = o I1/2r dF/dL = I B =  I I /2r

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dF/dL = I2B1 = o I1 I2 /2r

Force between Circular Loops

F 2 aI B Fz = 2aI2Br1

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Force between Circular Loops

Example: Two coaxial circular coils with a = 1 m, separated by z = 1 m. , p y Find F. k2 = 0 8 k = 0.8 K(k) = 2.257, E(k) = 1.178, Br = 0.0697 T F = 4.38x105 N

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Tensile Stress in Long Solenoid Coil

Example Case: Example Case: r1 = 1 m, ∆r = 0.2 m B = 10 T. = 212 MPa Yield stress of copper = 280 MPa

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Force on Torsatron Coils Force on Torsatron Coils

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Force Reduced Torsatron Coils

Optimum Pitch angle ~ 42o g R/ac ~ 7

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TF Coil Design Considerations g

TF coil forces tend to: i il di increase coil radius ac decrease major radius Rc bend coils (due to interaction with vertical field) Consider stress concentrations fatigue fatigue creep thermal stress TF coils shaped like “D” have lower stress than circular coils TF coils shaped like “D” have lower stress than circular coils.

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Reduction of Field Errors

Coil winding accuracy Coil alignment Coil supports to minimize motion Series connection to equalize currents Series connection to equalize currents Stray B fields from current leads Stray B fields from ferrous objects

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Components p

Energy storage Energy storage Switches Transmission lines Coils Diagnostics and controls Diagnostics and controls

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RLC Circuit Equations q

R = total resistance L t t l i d t L = total inductance L(d2q/dt2) +R(dq/dt)+q/C =0 L(d2q/dt2) +R(dq/dt)+q/C =0 q(0) = CVo (dq/dt)o = 0 q(t) = CVo e-at [cost + (a/)sint] I(t) = (Vo/L) e-at sint a=R/2L  = [(1/LC) a2]

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a=R/2L  = [(1/LC) – a2]

Current vs Time Current vs. Time

“Undercritically Undercritically damped circuit”

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“Crowbar” Switch S2

Close S1 at t=0 Close S2 at t=tmax

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Resistance of Wire, Rod, Plate, Tube Resistance of Wire, Rod, Plate, Tube

ℓ R  d / S R =  dx  / S Example: Copper tube r1 = 0.02 m, r2 = 0.025 m, ℓ = 3 m = 2x10-6 Ohm-m  2x10 Ohm m S = (r2

2-r1 2) = 0.000707 m2

R =  ℓ / S = 0.00849 Ohm

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Inductance of N-turn Solenoid Inductance of N turn Solenoid

Length ℓ , Radii r1 and r2  = r2/r1

1 2

Example: N=20, ℓ = 1 m r = 5 m ℓ = 1 m, r1 = .5 m, r2 = .8 m  = 1.6,  = 2, L/N2r1 = 1.2x10-6  = ℓ /r1 L = 2.4x10-4 Henry

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Parallel Plate Transmission Line Parallel Plate Transmission Line

L=oS ℓ Ksh / h

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Graph of K h vs (s/h) Graph of Ksh vs. (s/h)

If /h 1 If s/h << 1, then Ksh = 1

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Coaxial Cable or Tubes Coaxial Cable or Tubes

L =  ℓ ln(b/a)/2 L = o ℓ ln(b/a)/2

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Distribution of J and B Distribution of J and B

J/t = 2J/ J/t  J/ B/t = 2B/

Assume 2B ≈ B/2 B/t ≈ B Then B ≈ B/2   ≈  Ski d th  (2/ )1/2 Skin depth  = (2/)1/2 In copper at 1 MHz,  = 0.07 mm

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pp

Structural Support of Coil Structural Support of Coil

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Distribution of J and B in coil Distribution of J and B in coil

Actual Approximate

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Axial Distribution of B in Solenoids

Single-turn Uniform J

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Coil Melting and Yielding

Yielding at B ~ By[(r2-r1)/2r1]1/2 B (Cu SS) = 25 T B (Ta) = 32 T By (Cu, SS) = 25 T By(Ta) = 32 T Melting at B ~ B /1/2  ~ 3 Melting at B ~ Be/1/2  ~ 3 Be(Cu, SS) = 90 T Be(Ta) = 137 T Fatigue failures after many shots Sudden B > 70 T  Coil explodes

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Energy Storage Systems

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Scyllac Capacitor Scyllac Capacitor

60 kV, 1.85 F

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Energy Storage System Costs Energy Storage System Costs

Fusion experiments Power grids S l Solar power Wind power flywheel ~1980 values

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Inductive Energy Storage

Opening switch S1 forces current to flow through plasma confinement coil. S1 : difficult to prevent arcing

Transfer efficiency = LsL/(Ls+L)2  25%

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50 MJ Homopolar Generator

Er = vxBz

U of Texas

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• U. of Texas

Flywheel Energy Storage

Can store about 500 MJ/m3 Can store about 500 MJ/m Princeton Plasma Physics Laboratory Princeton Plasma Physics Laboratory 200 MW 3 motor-generator 200 MW, 3 s More expensive than homopolar system

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Spark Gap Switch p p

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Los Alamos Dual Spark Gap Switch

Low “jitter”: 3240 switches fired within 10 ns.

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Exploding Foil Switch

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Marx Bank

+400 kV Capacitors are charged in parallel 100 kV 100 kV 100 kV 100 kV Capacitors are charged in parallel Then discharged in series to give high voltage

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High Voltage Coaxial Cable

Scyllac experiment had 250 km of these cables Scyllac experiment had 250 km of these cables. 105 shot reliability.

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Magnetic Flux Compression

“Bellows” type

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Imploding Metallic Liner Imploding Metallic Liner

explosive liner Debris low flux Debris Liner Compressed flux

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Failure Rates

fjdt = failure probability of item j during dt t Failure probability between 0 and t: p =  dt f(t) Failure probability between 0 and t: pf =  dt fj(t) Probability of not failing before time t = 1 p Probability of not failing before time t = 1-pf “Failure Rate” at time t: rj(t) = pf/(1-pf) Total failure rate r(t) =  rj(t) (failures per second) j Estimated time between failures ETBF = 1/r(t)

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Component Reliability

Capacitors, cables, automobiles automobiles, …

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Estimated Time to Next Failure

Example: 100 capacitors with j = 10-4 per shot and 600 cables with j = 2x10-4 per shot Find ETNF Find ETNF ETNF = 1 / [100x10-4 + 600x2x10-4] = 7.7 shots Similar analysis for laser systems, automobiles, etc.

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Coil Power Requirements

= (copper volume)/(coil volume) dP = Jc

2dV

Circular coils: P =  Jc

2  dz  2rdr over coil volume

f Field at center of solenoid with length L, radii r1 and r2: Bz = 3/2 o g()(P/r1)1/2

z

o g( )(  1) where  = r2/r1  = L/2r1

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Relation of B to Input Power Relation of Bz to Input Power

Bz = 3/2 o g()(P/r1)1/2 Bz  o g()(P/r1)  = 2x10-8 Ohm-m Example:  = 0.9 r1 = 0.1 m, P = 100 kW Find optimum coil Bz Optimum g() =0.142 at Optimum g() 0.142 at r2 = 3r1, L = 4r1. B = 1 1 T

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Bz = 1.1 T

Coil Power Requirements Coil Power Requirements

Given r1 = 3 m, B = 10 T, find P Result: P = 240 MW Result: P 240 MW This is why big experiments use superconducting coils. Liquid N2 coolant (77 K) can lower  and required power.

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Heat Removal Rate

P =CmT(dV/dt)

C = specific heat of coolant  = mass density of coolant m mass density of coolant T = temperature rise of coolant dV/dt = volumetric flow rate of coolant = Awv A l t h l Aw = coolant channel area v = coolant flow speed

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Pumping Power p g

Reynold’s Number Re = dVm/ p = f Lcmv2/2D (Pa) Pumping power: Pc = p (dV/dt)/p p = pump efficiency

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Friction F t Factor

Re = dVm/

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Example – Pumping Power

100 kW coil, 16 coolant channels in parallel, each 30 m long, 4.6 mm diameter, t = 60 K. Find dV/dt v p P Find dV/dt, v, p, Pc Total dV/dt = P/CmT = 3.99x10-4 m3/s O h l (dV/d )/16 2 49 10 5

3/

One channel = (dV/dt)/16 = 2.49x10-5 m3/s Ac = 1.66x10-5 m2 v = (dV/dt)/A = 1.50 m/s ( ) Re = 6872 f = 0.035 from graph p = f L  v2/2D = 2 56x105 Pa p = f Lcmv /2D = 2.56x10 Pa Pc = p (dV/dt)/p = 128 W.

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Coil Winding Methods g

Hollow Conductor Tape-wound coil vz Conductor “Pancake coil”, v coil vz Bitter magnet vr

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Safe Value of Current Density Safe Value of Current Density

Heat dissipated in volume Vch cooled by one channel Pch = J2Vch Equate to heat removed by coolant: P =Cmt(dV/dt) Solve for safe value of J

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Maximum Safe Current

Assuming T = 60 K p 0 41 MPa p = 0.41 MPa Match coil resistance to power supply

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Coil Electrical Resistance

Rc = Lc/Ac Example: 16 coils in series, Example: 16 coils in series, each Lc = 30 m, Square copper with a = 8 64 mm D = 4 66 mm

D

a = 8.64 mm, D = 4.66 mm. Ac = a2 – D2/4 = 5.76x10-5 m2

a

One winding: Rc = 0.0104 Ohm 16 in Series R = 1.67 Ohm. Joints add resistance.

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Coil Winding Coil Winding

Wind around a form Hi h t i t b d t h High tension to bend copper to shape Accurate position of copper important Fiberglass insulation between layers Epoxy to hold conductor rigidly Brazed joints Brazed joints Aluminum joints unreliable

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Summary – Pulsed Magnets Summary Pulsed Magnets

Components and circuit calculations are simple, but diff i f J d B l diffusion of J and B are more complex. Capacitor banks are widely used for pulsed magnets y g Inductive storage and flywheels less expensive at very high energy very high energy Magnetic flux compression  very high B Component reliability for millions of shots is difficult

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Summary – Water-Cooled Magnets

Cryogenic insulation refrigeration not needed Cryogenic insulation, refrigeration not needed Brazed joints reliable Bolted joints easy to assemble and disassemble C f Can tolerate high neutron fluences Does not require stabilization Technology well developed, reliable gy p , Calculations of current, B field, coil power, and pumping power simple and reliable and pumping power simple and reliable.

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